3.6-bo'lim: Yechimlarning asosiy to'plami
Nihoyat, "etarlicha yoqimli" ni aniqlash vaqti keldi. Biz ushbu atamani so'nggi bir necha bo'limda umumiy yechimni shakllantirish uchun ishlatilishi mumkin bo'lgan echimlarni tasvirlash uchun ishlatdik va endi uni rasmiy ravishda aniqlash vaqti keldi.
Birinchidan, biz bu erda qiladigan hamma narsa faqat chiziqli va bir hillikni talab qilganligi sababli, biz differentsial tenglamada doimiy koeffitsientlarni talab qilmaymiz. Shunday qilib, keling, quyidagi IVP bilan boshlaylik.
\[\begin{equation}\begin{array}{c}p\left( t \right)y'' + q\left( t \right)y' + r\left( t \right)y = 0\\ y\left( {{t_0}} \right) = {y_0}\hspace{0.25in}\,\,y'\left( {{t_0}} \right) = {{y'}_0}\end{array}\label{eq:eq1}\end{equation}\]Aytaylik, biz ushbu differentsial tenglamaning ikkita yechimini topdik, \(y_{1}(t)\) va \(y_{2}(t)\). dan bilamiz Superpozitsiya printsipi bu
\[\begin{equation}y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)\label{eq:eq2}\end{equation}\]differensial tenglamaning yechimi ham bo'ladi. Biz bilmoqchi bo'lgan narsa bu umumiy yechim bo'ladimi yoki yo'qmi. \(\eqref{eq:eq2}\) umumiy yechim deb hisoblanishi uchun u \(\eqref{eq:eq1}\) da umumiy boshlang‘ich shartlarga javob berishi kerak.
\[y\left( {{t_0}} \right) = {y_0}\hspace{0.25in}\,\,\,\,\,\,\,y'\left( {{t_0}} \right) = {y'_0}\]Bu shuningdek, differensial tenglamaning har qanday yechimini ushbu shaklda yozish mumkinligini anglatadi.
Shunday qilib, keling, ushbu shartlarni qondiradigan doimiylarni topa olamizmi yoki yo'qligini bilib olaylik. Avval \(\eqref{eq:eq2}\) ni ajratib oling va boshlang'ich sharoitlarni ulang.
\[\begin{equation}\begin{array}{l}{y_0} = y\left( {{t_0}} \right) = {c_1}{y_1}\left( {{t_0}} \right) + {c_2}{y_2}\left( {{t_0}} \right)\\ {{y'}_0} = y'\left( {{t_0}} \right) = {c_1}{{y'}_1}\left( {{t_0}} \right) + {c_2}{{y'}_2}\left( {{t_0}} \right)\end{array}\label{eq:eq3}\end{equation}\]Bizda ikkita yechim bor deb taxmin qilganimiz sababli, ushbu tizimdagi hamma narsa texnik jihatdan ma'lum va shuning uchun bu \(c_{1}\) va \(c_{2}\) uchun echilishi mumkin bo'lgan tizim. Buni umuman Kramer qoidasi yordamida amalga oshirish mumkin. Kramer qoidasidan foydalanish quyidagi yechimni beradi.
\[\begin{equation}{c_1} = \frac{{\left| {\begin{array}{*{20}{c}}{{y_0}}&{{y_2}\left( {{t_0}} \right)}\\{{{y'}_0}}&{{{y'}_2}\left( {{t_0}} \right)}\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}{{y_1}\left( {{t_0}} \right)}&{{y_2}\left( {{t_0}} \right)}\\{{{y'}_1}\left( {{t_0}} \right)}&{{{y'}_2}\left( {{t_0}} \right)}\end{array}} \right|}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{c_2} = \frac{{\left| {\begin{array}{*{20}{c}}{{y_1}\left( {{t_0}} \right)}&{{y_0}}\\{{{y'}_1}\left( {{t_0}} \right)}&{{{y'}_0}}\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}{{y_1}\left( {{t_0}} \right)}&{{y_2}\left( {{t_0}} \right)}\\{{{y'}_1}\left( {{t_0}} \right)}&{{{y'}_2}\left( {{t_0}} \right)}\end{array}} \right|}}\label{eq:eq4}\end{equation}\]qayerda,
\[\left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right| = ad - bc\]2x2 matritsaning determinantidir. Agar siz aniqlovchilar haqida bilmasangiz, yuqorida keltirilgan formuladan foydalaning.
Endi \(\eqref{eq:eq4}\) \(\eqref{eq:eq3}\) tizimiga yechim beradi. E'tibor bering, amalda biz tizimlarni echish uchun odatda Kramer qoidasidan foydalanmaymiz, biz oddiy yo'l bilan harakat qilamiz va asosiy algebra texnikasi yordamida tizimni yechamiz. Xo'sh, nega biz bu erda Kramer qoidasidan foydalandik?
Biz Kramer qoidasidan foydalandik, chunki biz \(\eqref{eq:eq4}\) dan konstantalarni qachon yechishimizni aniqlashga imkon beradigan shartni ishlab chiqishimiz mumkin. \(\eqref{eq:eq4}\) dagi miqdorlarning uchtasi (ha, uchtasi, maxrajlari bir xil!) shunchaki raqamlar va maxraj nolga teng bo'lganda, haqiqatda yechim topishimizga to'sqinlik qiladigan yagona narsa bo'ladi.
Maxrajdagi miqdor deyiladi Vronskiy va sifatida belgilanadi
\[W\left( {f,g} \right)\left( t \right) = \left| {\begin{array}{*{20}{c}}{f\left( t \right)}&{g\left( t \right)}\\{f'\left( t \right)}&{g'\left( t \right)}\end{array}} \right| = f\left( t \right)g'\left( t \right) - g\left( t \right)f'\left( t \right)\]Funktsiyalar va/yoki \(t\) nima ekanligi aniq bo'lsa, biz ko'pincha Wronskianni \(W\) bilan belgilaymiz.
Keling, bu erda nimadan keyin bo'lganimizni eslaylik. Biz \(\eqref{eq:eq1}\) ga ikkita yechim qachon umumiy yechim yaratish uchun yetarli bo‘lishini aniqlamoqchi edik. Ikki yechim \(\eqref{eq:eq1}\) ning umumiy yechimini hosil qiladi, agar ular \(\eqref{eq:eq1}\) da berilgan umumiy boshlang‘ich shartlarga javob bersa va biz Kramer qoidasidan Vronskian nolga teng bo‘lmasa, ular dastlabki shartlarga javob berishini ko‘rishimiz mumkin. Yoki,
\[W\left( {{y_1},{y_2}} \right)\left( {{t_0}} \right) = \left| {\begin{array}{*{20}{c}}{{y_1}\left( {{t_0}} \right)}&{{y_2}\left( {{t_0}} \right)}\\{{{y'}_1}\left( {{t_0}} \right)}&{{{y'}_2}\left( {{t_0}} \right)}\end{array}} \right| = {y_1}\left( {{t_0}} \right){y'_2}\left( {{t_0}} \right) - {y_2}\left( {{t_0}} \right){y'_1}\left( {{t_0}} \right) \ne 0\]Shunday qilib, \(y_{1}(t)\) va \(y_{2}(t)\) \(\eqref{eq:eq1}\) va \(W\left( {{y_1},{y_2}} \right)\left( t \right) \ne 0\) ning ikkita yechimi deb faraz qilaylik. Keyin ikkita yechim a deyiladi asosiy echimlar to'plami va \(\eqref{eq:eq1}\) ning umumiy yechimi
\[y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right)\]Endi biz "etarlicha yaxshi" nimani anglatishini bilamiz. Ikkita yechim, agar ular asosiy echimlar to'plami bo'lsa, "etarlicha yoqimli".
Shunday qilib, keling, oldingi bo'limda qilgan da'volardan birini tekshirib ko'raylik. Qolgan ikkitasini xohlaysizmi, tekshirish uchun sizga qoldiramiz.
yechimlarning asosiy to‘plami edi. Ular haqiqatan ham borligini isbotlang.
Yechim
Shunday qilib, buni isbotlash uchun biz ushbu ikkita yechim uchun Wronskianni olishimiz va uning nolga teng emasligini ko'rsatishimiz kerak.
\[\begin{align*}W & = \left| {\begin{array}{*{20}{c}}{{{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right)}&{{{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)}\\{\lambda {{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right) - \mu {{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)}&{\lambda {{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right) + \mu {{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right)}\end{array}} \right|\\ & = {{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right)\left( {\lambda {{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right) + \mu {{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right)} \right) - \\ & \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)\left( {\lambda {{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right) - \mu {{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)} \right)\\ & = \mu {{\bf{e}}^{2\lambda t}}{\cos ^2}\left( {\mu \,t} \right) + \mu {{\bf{e}}^{2\lambda t}}{\sin ^2}\left( {\mu \,t} \right)\\ & = \mu {{\bf{e}}^{2\lambda t}}\left( {{{\cos }^2}\left( {\mu \,t} \right) + {{\sin }^2}\left( {\mu \,t} \right)} \right)\\ & = \mu {{\bf{e}}^{2\lambda t}}\end{align*}\]Endi eksponensial hech qachon nolga teng bo'lmaydi va \(\mu \ne 0\) (agar bizda murakkab ildizlar bo'lmasa edi!) va shuning uchun \(W \ne 0\). Shuning uchun, bu ikki yechim aslida yechimlarning asosiy to'plamidir va shuning uchun bu holda umumiy yechim.
\[y\left( t \right) = {c_1}{{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right) + {c_2}{{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)\]Ushbu ikkinchi yechim berilgan yechim bilan birga differensial tenglama uchun fundamental yechimlar to‘plamini tashkil qilishini ko‘rsating.
Yechim
Ushbu misoldagi ikkita yechim
\[{y_1}\left( t \right) = {t^{ - 1}}\hspace{0.25in}\hspace{0.25in}{y_2}\left( t \right) = {t^{\frac{3}{2}}}\]Keling, ushbu ikkita yechimning Wronskianini hisoblaymiz.
\[W = \left| {\begin{array}{*{20}{c}}{{t^{ - 1}}}&{{t^{\frac{3}{2}}}}\\{ - {t^{ - 2}}}&{\frac{3}{2}{t^{\frac{1}{2}}}}\end{array}} \right| = \frac{3}{2}{t^{ - \,\frac{1}{2}}} - \left( { - {t^{ - \,\frac{1}{2}}}} \right) = \frac{5}{2}{t^{ - \,\frac{1}{2}}} = \frac{5}{{2\sqrt t }}\]Shunday qilib, Wronskian hech qachon nolga teng bo'lmaydi. E'tibor bering, biz \(t\) = 0 ni Wronskian-ga ulay olmaymiz. Bu umumiy yechimdagi konstantalarni topishda muammo bo'ladi, bundan tashqari biz \(t\) = 0 ni ham yechimga ulay olmaymiz va shuning uchun bu muammo bo'lib tuyulishi mumkin emas.
Shunday qilib, har qanday \(t\) uchun Wronskian nolga teng bo'lmaganligi sababli, ikkita echim asosiy echimlar to'plamini tashkil qiladi va umumiy yechim
\[y\left( t \right) = {c_1}{t^{ - 1}} + {c_2}{t^{\frac{3}{2}}}\]biz o'sha misolda da'vo qilganimizdek.
Shu nuqtaga qadar biz yechimlar to'plamini topdik, keyin ular aslida asosiy echimlar to'plami ekanligini da'vo qildik. Albatta, siz hozir biz bildirgan barcha da'volarni tekshirishingiz mumkin, ammo bu savol tug'diradi. Berilgan differensial tenglama uchun fundamental yechimlar to‘plami mavjudligini qanday bilamiz? Bu savolga quyidagi teorema javob beradi.
Yechimlarning asosiy to'plamining mavjudligi
Differensial tenglamani ko'rib chiqing
\[y'' + p\left( t \right)y' + q\left( t \right)y = 0\]bu yerda \(p(t)\) va \(q(t)\) ba'zi I oraliqda uzluksiz funksiyalardir. I oraliqdagi istalgan nuqta bo'lishi uchun \(t_{0}\) ni tanlang. \(y_{1}(t)\) differensial tenglamaning boshlang'ich shartlarini qanoatlantiradigan yechimi bo'lsin.
\[y\left( {{t_0}} \right) = 1\hspace{0.25in}y'\left( {{t_0}} \right) = 0\]\(y_{2}(t)\) differensial tenglamaning boshlang‘ich shartlarini qanoatlantiruvchi yechimi bo‘lsin.
\[y\left( {{t_0}} \right) = 0\hspace{0.25in}y'\left( {{t_0}} \right) = 1\]Keyin \(y_{1}(t)\) va \(y_{2}(t)\) differentsial tenglama uchun asosiy yechimlar to'plamini tashkil qiladi.
Bu ikki yechim yechimlarning asosiy to'plamini tashkil etishini ko'rsatish juda oson. Faqat Wronskianni hisoblang.
\[W\left( {{y_1},{y_2}} \right)\left( {{t_0}} \right) = \left| {\begin{array}{*{20}{c}}{{y_1}\left( {{t_0}} \right)}&{{y_2}\left( {{t_0}} \right)}\\{{{y'}_1}\left( {{t_0}} \right)}&{{{y'}_2}\left( {{t_0}} \right)}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right| = 1 - 0 = 1 \ne 0\]Shunday qilib, biz teoremada berilgan ikkita IVP ni hal qila olsak, asosiy echimlar to'plami mavjud bo'ladi.
\(t_{0} = 0\) yordamida.
Yechim
Ushbu bobning birinchi qismidagi usullardan foydalanib, biz hozirgacha ishlatib kelgan ikkita yechimni topishimiz mumkin.
\[y\left( t \right) = {{\bf{e}}^{ - 3t}}\hspace{0.25in}\hspace{0.25in}y\left( t \right) = {{\bf{e}}^{ - t}}\]Bu yechimlarning asosiy to'plamini tashkil qiladi, chunki biz osongina tekshirishimiz mumkin. Biroq, ular teorema tomonidan beriladigan to'plam emas. Bu yechimlarning hech biri teoremada berilgan boshlang‘ich shartlarning ikkita to‘plamini ham qanoatlantirmaydi. Teorema tomonidan berilgan asosiy echimlar to'plamini topish uchun biz ulardan foydalanishimiz kerak.
Biz bilamizki, quyida ham differensial tenglamaning yechimi hisoblanadi.
\[y\left( t \right) = {c_1}{{\bf{e}}^{ - 3t}} + {c_2}{{\bf{e}}^{ - t}}\]Shunday qilib, keling, dastlabki shartlarning birinchi to'plamini qo'llaymiz va biz ishlaydigan doimiylarni topa olamizmi yoki yo'qligini bilib olaylik.
\[y\left( 0 \right) = 1\hspace{0.25in}y'\left( 0 \right) = 0\]Buni amalga oshirgandan so'ng quyidagi yechimga ega ekanligimizni tekshirishni sizga qoldiramiz.
\[{y_1}\left( t \right) = - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{3}{2}{{\bf{e}}^{ - t}}\]Xuddi shunday, agar biz dastlabki shartlarning ikkinchi to'plamini qo'llasak,
\[y\left( 0 \right) = 0\hspace{0.25in}y'\left( 0 \right) = 1\]olamiz
\[{y_2}\left( t \right) = - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{1}{2}{{\bf{e}}^{ - t}}\]Teoremaga ko'ra, ular yechimlarning asosiy to'plamini tashkil qilishi kerak. Buni tekshirish juda oson.
\[\begin{align*}W & = \left| {\begin{array}{*{20}{c}}{ - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{3}{2}{{\bf{e}}^{ - t}}}&{ - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{1}{2}{{\bf{e}}^{ - t}}}\\{\frac{3}{2}{{\bf{e}}^{ - 3t}} - \frac{3}{2}{{\bf{e}}^{ - t}}}&{\frac{3}{2}{{\bf{e}}^{ - 3t}} - \frac{1}{2}{{\bf{e}}^{ - t}}}\end{array}} \right|\\ & = \left( { - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{3}{2}{{\bf{e}}^{ - t}}} \right)\left( {\frac{3}{2}{{\bf{e}}^{ - 3t}} - \frac{1}{2}{{\bf{e}}^{ - t}}} \right) - \left( { - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{1}{2}{{\bf{e}}^{ - t}}} \right)\left( {\frac{3}{2}{{\bf{e}}^{ - 3t}} - \frac{3}{2}{{\bf{e}}^{ - t}}} \right)\\ & = {{\bf{e}}^{ - 4t}} \ne 0\end{align*}\]Shunday qilib, biz teoremadan shu paytgacha ishlatganimizdan butunlay boshqacha asosiy echimlar to'plamini oldik. Bu muammo emas. Ushbu muammoni hal qilishning asosiy to'plami sifatida foydalanishimiz mumkin bo'lgan cheksiz ko'p funktsiyalar juftligi mavjud.
Xo'sh, qaysi asosiy echimlardan foydalanishimiz kerak? Xo'sh, agar biz dastlab topganlarimizdan foydalansak, umumiy yechim shunday bo'ladi:
\[y\left( t \right) = {c_1}{{\bf{e}}^{ - 3t}} + {c_2}{{\bf{e}}^{ - t}}\]Holbuki, agar biz teoremadan to'plamdan foydalansak, umumiy yechim shunday bo'ladi:
\[y\left( t \right) = {c_1}\left( { - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{3}{2}{{\bf{e}}^{ - t}}} \right) + {c_2}\left( { - \frac{1}{2}{{\bf{e}}^{ - 3t}} + \frac{1}{2}{{\bf{e}}^{ - t}}} \right)\]Boshlang'ich shartlarning umumiy to'plamini qondirish uchun koeffitsientlarni aniqlashga kelganda, bu bilan ishlash juda qiziqarli bo'lmaydi.
Xo'sh, qaysi asosiy echimlardan foydalanishimiz kerak? Biz har doim berilgan muammo uchun foydalanish uchun eng qulay bo'lgan to'plamdan foydalanishga harakat qilishimiz kerak.